3.464 \(\int \frac {1}{(d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}})^{3/2}} \, dx\)
Optimal. Leaf size=158 \[ \frac {3 a f^2 \tanh ^{-1}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{2 d^{5/2} e}-\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^2 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}-\frac {\frac {a f^2}{d^2}+1}{e \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}} \]
[Out]
3/2*a*f^2*arctanh((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^(1/2))/d^(5/2)/e+(-1-a*f^2/d^2)/e/(d+e*x+f*(a+e^2*x^
2/f^2)^(1/2))^(1/2)-1/2*a*f^2*(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^2/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))
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Rubi [A] time = 0.16, antiderivative size = 158, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used =
{2117, 897, 1259, 453, 206} \[ -\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^2 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}-\frac {\frac {a f^2}{d^2}+1}{e \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}+\frac {3 a f^2 \tanh ^{-1}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{2 d^{5/2} e} \]
Antiderivative was successfully verified.
[In]
Int[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-3/2),x]
[Out]
-((1 + (a*f^2)/d^2)/(e*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])) - (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2
)/f^2]])/(2*d^2*e*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) + (3*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^
2]]/Sqrt[d]])/(2*d^(5/2)*e)
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 453
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Rule 897
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]
Rule 1259
Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Rule 2117
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
Rubi steps
\begin {align*} \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {d^2+a f^2-2 d x+x^2}{(d-x)^2 x^{3/2}} \, dx,x,d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {d^2+a f^2-2 d x^2+x^4}{x^2 \left (d-x^2\right )^2} \, dx,x,\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}\right )}{e}\\ &=-\frac {a f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{2 d^2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 d \left (d^2+a f^2\right )+\left (2 d^2-a f^2\right ) x^2}{x^2 \left (d-x^2\right )} \, dx,x,\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}\right )}{2 d^2 e}\\ &=-\frac {d^2+a f^2}{d^2 e \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}-\frac {a f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{2 d^2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {\left (3 a f^2\right ) \operatorname {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}\right )}{2 d^2 e}\\ &=-\frac {d^2+a f^2}{d^2 e \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}-\frac {a f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{2 d^2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {3 a f^2 \tanh ^{-1}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{2 d^{5/2} e}\\ \end {align*}
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Mathematica [A] time = 0.43, size = 167, normalized size = 1.06 \[ \frac {\frac {3 a f^2 \tanh ^{-1}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{d^{5/2}}+\frac {-2 d^2 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )-a f^2 \left (3 f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+3 e x\right )}{d^2 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right ) \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}}{2 e} \]
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-3/2),x]
[Out]
((-2*d^2*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) - a*f^2*(d + 3*e*x + 3*f*Sqrt[a + (e^2*x^2)/f^2]))/(d^2*(e*x + f*Sq
rt[a + (e^2*x^2)/f^2])*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]) + (3*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a +
(e^2*x^2)/f^2]]/Sqrt[d]])/d^(5/2))/(2*e)
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fricas [A] time = 0.52, size = 487, normalized size = 3.08 \[ \left [\frac {3 \, {\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \sqrt {d} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} - 2 \, {\left (\sqrt {d} e x - \sqrt {d} f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) - 2 \, {\left (2 \, d^{2} e^{2} x^{2} - 2 \, a d^{2} f^{2} - 2 \, d^{4} - {\left (3 \, a d e f^{2} + d^{3} e\right )} x + {\left (3 \, a d f^{3} - 2 \, d^{2} e f x + d^{3} f\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{4 \, {\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}}, -\frac {3 \, {\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt {-d}}{d}\right ) + {\left (2 \, d^{2} e^{2} x^{2} - 2 \, a d^{2} f^{2} - 2 \, d^{4} - {\left (3 \, a d e f^{2} + d^{3} e\right )} x + {\left (3 \, a d f^{3} - 2 \, d^{2} e f x + d^{3} f\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{2 \, {\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="fricas")
[Out]
[1/4*(3*(a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^2)*sqrt(d)*log(a*f^2 - 2*d*e*x + 2*d*f*sqrt((e^2*x^2 + a*f^2)/f^2)
- 2*(sqrt(d)*e*x - sqrt(d)*f*sqrt((e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)) - 2*(
2*d^2*e^2*x^2 - 2*a*d^2*f^2 - 2*d^4 - (3*a*d*e*f^2 + d^3*e)*x + (3*a*d*f^3 - 2*d^2*e*f*x + d^3*f)*sqrt((e^2*x^
2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d))/(a*d^3*e*f^2 - 2*d^4*e^2*x - d^5*e), -1/2*(3*(
a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^2)*sqrt(-d)*arctan(sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)*sqrt(-d)/d)
+ (2*d^2*e^2*x^2 - 2*a*d^2*f^2 - 2*d^4 - (3*a*d*e*f^2 + d^3*e)*x + (3*a*d*f^3 - 2*d^2*e*f*x + d^3*f)*sqrt((e^
2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d))/(a*d^3*e*f^2 - 2*d^4*e^2*x - d^5*e)]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const ge
n & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueWarning, choosing root of [1,0,%%%{-2,[1,2
,0,0]%%%}+%%%{-2,[0,1,0,0]%%%}+%%%{-2,[0,0,2,1]%%%},0,%%%{1,[2,4,0,0]%%%}+%%%{-2,[1,3,0,0]%%%}+%%%{2,[1,2,2,1]
%%%}+%%%{1,[0,2,0,0]%%%}+%%%{-2,[0,1,2,1]%%%}+%%%{1,[0,0,4,2]%%%}] at parameters values [-49,85.3561567818,-64
,-30]Warning, choosing root of [1,0,%%%{-2,[1,2,0,0]%%%}+%%%{-2,[0,1,0,0]%%%}+%%%{-2,[0,0,2,1]%%%},0,%%%{1,[2,
4,0,0]%%%}+%%%{-2,[1,3,0,0]%%%}+%%%{2,[1,2,2,1]%%%}+%%%{1,[0,2,0,0]%%%}+%%%{-2,[0,1,2,1]%%%}+%%%{1,[0,0,4,2]%%
%}] at parameters values [0,61.7937478349,0,0]Warning, choosing root of [1,0,%%%{-2,[1,2,0,0]%%%}+%%%{-2,[0,1,
0,0]%%%}+%%%{-2,[0,0,2,1]%%%},0,%%%{1,[2,4,0,0]%%%}+%%%{-2,[1,3,0,0]%%%}+%%%{2,[1,2,2,1]%%%}+%%%{1,[0,2,0,0]%%
%}+%%%{-2,[0,1,2,1]%%%}+%%%{1,[0,0,4,2]%%%}] at parameters values [56,62.4600259969,-13,46]Warning, choosing r
oot of [1,0,%%%{-2,[1,2,0,0]%%%}+%%%{-2,[0,1,0,0]%%%}+%%%{-2,[0,0,2,1]%%%},0,%%%{1,[2,4,0,0]%%%}+%%%{-2,[1,3,0
,0]%%%}+%%%{2,[1,2,2,1]%%%}+%%%{1,[0,2,0,0]%%%}+%%%{-2,[0,1,2,1]%%%}+%%%{1,[0,0,4,2]%%%}] at parameters values
[-6,25.8388736797,81,18]Warning, choosing root of [1,0,%%%{-2,[1,2,0,0]%%%}+%%%{-2,[0,1,0,0]%%%}+%%%{-2,[0,0,
2,1]%%%},0,%%%{1,[2,4,0,0]%%%}+%%%{-2,[1,3,0,0]%%%}+%%%{2,[1,2,2,1]%%%}+%%%{1,[0,2,0,0]%%%}+%%%{-2,[0,1,2,1]%%
%}+%%%{1,[0,0,4,2]%%%}] at parameters values [63,31.8503101398,2,62]Warning, choosing root of [1,0,%%%{-2,[1,2
,0,0]%%%}+%%%{-2,[0,1,0,0]%%%}+%%%{-2,[0,0,2,1]%%%},0,%%%{1,[2,4,0,0]%%%}+%%%{-2,[1,3,0,0]%%%}+%%%{2,[1,2,2,1]
%%%}+%%%{1,[0,2,0,0]%%%}+%%%{-2,[0,1,2,1]%%%}+%%%{1,[0,0,4,2]%%%}] at parameters values [0,10.4309062702,0,0]W
arning, choosing root of [1,0,%%%{-2,[1,2,0,0]%%%}+%%%{-2,[0,1,0,0]%%%}+%%%{-2,[0,0,2,1]%%%},0,%%%{1,[2,4,0,0]
%%%}+%%%{-2,[1,3,0,0]%%%}+%%%{2,[1,2,2,1]%%%}+%%%{1,[0,2,0,0]%%%}+%%%{-2,[0,1,2,1]%%%}+%%%{1,[0,0,4,2]%%%}] at
parameters values [65,39.1803401988,28,-44]Warning, choosing root of [1,0,%%%{-2,[1,2,0,0]%%%}+%%%{-2,[0,1,0,
0]%%%}+%%%{-2,[0,0,2,1]%%%},0,%%%{1,[2,4,0,0]%%%}+%%%{-2,[1,3,0,0]%%%}+%%%{2,[1,2,2,1]%%%}+%%%{1,[0,2,0,0]%%%}
+%%%{-2,[0,1,2,1]%%%}+%%%{1,[0,0,4,2]%%%}] at parameters values [91,88.2886286299,-21,88]Evaluation time: 0.64
Done
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maple [F] time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e x +d +\sqrt {\frac {e^{2} x^{2}}{f^{2}}+a}\, f \right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(e*x+d+(e^2/f^2*x^2+a)^(1/2)*f)^(3/2),x)
[Out]
int(1/(e*x+d+(e^2/f^2*x^2+a)^(1/2)*f)^(3/2),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="maxima")
[Out]
integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(-3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^(3/2),x)
[Out]
int(1/(d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+e**2*x**2/f**2)**(1/2))**(3/2),x)
[Out]
Integral((d + e*x + f*sqrt(a + e**2*x**2/f**2))**(-3/2), x)
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